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4.
m?,
CHAPTER
1
Vectors
' 1L
1
Review of Prerequisite Skills ... 2
Rich Learning Link ... 4
1.1
Vector Concepts ... 5
1.2
Vector Laws ... 11
1.3
Force as a Vector... 18
1.4
Velocity as a Vector... 27
Key Concepts Review ... 33
Rich Learning Link WrapUp ... 34
Review Exercise ... 35
Chapter 1 Test... 37
Extending and Investigating ... 38
CHAPTER 2
Algebraic Vectors and Applications
41
Rich Learning Link ... 42
2.1
2.2
2.3
2.4
2.5
Algebraic Vectors ... 43
Operations with Algebraic Vectors ... 52
The Dot Product of Two Vectors ... 57
The Cross Product of Two Vectors ... 63
Applications of Dot and Cross Products ... 69
Review Exercise ... 76
Rich Learning Link WrapUp ... 78
Chapter 2 Test... 79
Extending and Investigating ... 81
CHAPTER 3
Lines in a Plane
83
Review of Prerequisite Skills ... 84
Rich Learning Link ... 85
3.1
Parametric and Vector Equations of a Line in a Plane ... 86
3.2
3.3
The Scalar Equation of a Line in a Plane ... 94
Equations of a Line in 3Space ... 99
The Intersection of Two Lines ... 105
3.4
Key Concepts Review ... 111
Review Exercise ... 112
Rich Learning Link WrapUp ... 115
Chapter 3 Test... 116
Extending and Investigating ... 117
CONTENTS
iii
5.
CHAPTER 4
Equations of Planes
119
Review of Prerequisite Skills ... 120
Rich Learning Link ... 121
4.1
4.2
4.3
4.4
4.5
The
The
The
The
The
Key
Vector Equation of a Plane in Space ... 122
Scalar Equation of a Plane in Space ... 128
Intersection of a Line and a Plane ... 135
Intersection of Two Planes ... 141
Intersection of Three Planes ... 148
Concepts Review ... 156
Rich Learning Link WrapUp ... 157
Review Exercise ... 158
Chapter 4 Test... 161
Extending and Investigating ... 163
Cumulative Review Chapters 14 ... 165
Glossary ... 169
Answers ... 173
Index ... 177
iv
CONTENTS
6.
T>
r
Have you ever tried to swim across a river with a
CHAPTER EXPECTATIONS In this chapter, you will
strong current? Have you sailed a boat, or run into
o represent vectors as directed line segments,
a head wind? If your answer is yes, then you have
experienced the effect of vector quantities. Vectors
were developed in the middle of the nineteenth
century as mathematical tools for studying physics.
In the following century, vectors became an
essential tool of navigators, engineers, and
physicists. In order to navigate, pilots need to know
what effect a crosswind will have on the direction
in which they intend to fly. In order to build
bridges, engineers need to know what load a
particular design will support. Physicists use vectors
in determining the thrust required to move a space
shuttle in a certain direction. You will learn more
about vectors in this chapter, and how vectors
represent quantities possessing both magnitude
and direction.
Section 1.1
o determine the components and projection of a
geometric vector. Section 1.1
o perform mathematical operations on geometric
vectors. Section 1.2
o model and.solve problems involving velocity and
force, Section 1.3, 1.4
7.
A vector is a quantity, an inseparable part of which is a direction. Pause for
a moment and think about physical quantities that have a direction. Force is an
example. The force of gravity acts only downward, never sideways. Wind is
another example. A wind from the north and a wind from the south have different
physical consequences, even if the wind speeds are the same. Temperature, on the
other hand, is not a vector quantity. Temperature does not go in any direction.
Temperature is referred to as a scalar quantity.
We need both scalar and vector quantities to model complex physical systems.
Meteorologists, for example, need data on air temperature and wind velocity,
among other things, to make weather forecasts.
The object of this chapter is to introduce the mathematical properties of vectors
and then show how vectors and scalars are used to describe many features of the
physical world.
In this chapter, we introduce the concept of a vector, a mathematical object
representing a physical quantity that has both magnitude and direction. We
will concentrate on geometric representations of vectors, so that most of our
discussion will be of twodimensional vectors. In later chapters we will introduce
algebraic representations of vectors, which will be more easily extended to higher
dimensions.
Before we begin this chapter, we will review some basic facts of trigonometry.
TRIGONOMETRIC RATIOS
In a rightangled triangle, as shown,
sin 8 = 
cos 8 = c
tan 6 = x
b
Note: The ratios depend on which angle is 6 and which
angle is 90°.
CHAPTER 1
8.
THE SINE LAW
a
_
sin A
b
_
sin B
c
sin C
THE COSINE LAW
a2 = b2 + c2 — 2bc cos A or cos A =
b2 + c2  a2
1. State the exact value of each of the following.
a. sin 60°
b. cos 60°
c. sin 135°
d. tan 120°
e. cos 30°
f.
tan 45°
2. A triangle ABC has AB = 6, ZB = 90°, and AC = 10. State the exact value
of tanA.
3. In AXYZ, XY = 6, ZX = 60°, and ZY = 70°. Determine the values of XZ, YZ,
and ZZ to twodecimal accuracy.
4. In &PQR, PQ = 4, PR = 1, and Qfl = 5. Determine the measures of the
angles to the nearest degree.
5. An aircraft control tower Tis tracking two planes at points A, 3.5 km from T,
and B, 6 km from T. If ZATB = 70°, determine the distance between the
planes.
6. Three ships are at points A, B, and C such that AB = 2 km, AC = 1 km,
and ZBAC = 142°. What is the distance between B and C?
REVIEW OF PREREQUISITE SKILLS
3
9.
IT ■■:*■■■";■ ■' :"1 >,.;■..' i . :. .;■ . ■'. ■'
CHAPTER 1: VECTORS AND*HE SUPERIOR COLLICULUS
Neuroscientists have found cells in a deep layer of a part of the brain called the
superior colliculus. These cells are tuned to the directions of distant visual and
auditory stimuli. Each cell responds only to stimuli from a specific direction.
Different cells are tuned to different directions. The tuning is broad, and the
regions to which different cells are tuned overlap considerably. Neuroscientists
have asked what it is about the activity in a group of cells with overlapping tuning
regions that specifies the actual direction of a stimulus. For example, how is it
that we can point accurately in the direction of a distant sound without
seeing its source? One answer is that a cell responds more vigorously when the
distance stimulus is in its direction. The direction is determined not by which cell
fires most vigorously, but by a type of addition of the degrees to which the
various cells have responded to the stimulus.
Investigate and Inquire
The type of addition performed in the brain can be illustrated by
a simple case involving only two brain cells. Suppose that one of
these cells responds to stimuli that are approximately north,
while the other responds to stimuli that are approximately east.
If the north cell responds twice as vigorously as the east cell,
what is the direction of the stimulus? We can use vector addition
to find out.
The answer is found by forming a triangle
with a side pointing east and a side
pointing north. The side pointing north
is twice as long as the side pointing east.
The third side is the actual direction of
the stimulus. From the diagram, we see
east
1 unit
north
2 units
direction
of stimulus
tan 6 = j. Solving, we find
9 = tan1 () s 26.6°.
So 6 = 26.6°.
Thus, the stimulus is 26.6° east of north.
What direction would be represented by a northeast cell
responding three times as vigorously as an east cell?
DISCUSSION QUESTIONS
1. How many cells would be needed to represent all the directions in the plane?
2. Why do you think the direction is not just taken to be the one corresponding
to the cell that fires most vigorously? #
CHAPTER 1
10.
Vectors are a part of everyone's common experience. Consider a typical winter
weather report that you might hear on the nightly news: The temperature is
presently 11 °C, with windfrom the northwest at 22 bn/lu This weather report
contains two different types of quantities. One quantity (the temperature)
is expressed as a single numerical value. The other quantity (the wind velocity)
has a numerical value (its magnitude) and also a direction associated with it.
These quantities are typical of the kinds encountered in science. They are
classified as follows:
Quantities having magnitude only are called scalars.
Quantities having both magnitude and direction are called vectors.
There seems to be some overlap here. For example, the temperature could be
thought of as having magnitude (11°) and direction (negative); in that sense, it
could be considered as a onedimensional vector. There is no problem with this
interpretation; sometimes it is a useful way to look at such quantities. However, in
most situations we find it easier to use positive and negative numbers as scalars,
and restrict the term vectors to quantities that require (at least) two properties to
define them.
Some examples of vector quantities are:
Force
The force of gravity has a well defined magnitude and acts in
a specific direction (down). The force of gravity is measured
when you step on a scale. Force is a vector quantity.
Displacement
When you walk from point A to point B, you travel a certain
distance in a certain direction. Displacement is a vector
quantity.
Magnetic Field
Some magnets are strong; others are weak. All cause
a compass needle to swing around and point in a particular
direction. A magnetic field is a vector quantity.
In a diagram, a vector is represented by an arrow: ^^r. The length of the arrow
is a positive real number and represents the magnitude of the vector. The direction
in which the arrow points is the direction of the vector. For now we will restrict
our discussion to vectors in two dimensions or to situations that can be expressed
in two dimensions. Our definitions and conclusions are easily extended to three
dimensions (or more).
1.1 VECTOR CONCEPTS
11.
EXAMPLE 1
A student travels to school by bus, first riding 2 km west, then changing buses and
riding a further 3 km north. Represent these displacements on
a vector diagram.
school
N
D
w
Solution
Suppose you represent a 1km distance by a 1cm line
segment. Then, a 2cm arrow pointing left represents the first
leg of the bus trip. A 3cm arrow pointing up represents the
bus
home
stop
second leg. The total trip is represented by a diagram
combining these vectors.
The notation used to describe vector quantities is as follows:
The algebraic symbol used in this text for a vector
is a letter with an arrow on top. Some texts use
m, v are vectors
boldface letters for vectors.
u, v are also vectors
Scalar quantities are written as usual.
x, y, a, b are scalars
The magnitude of a vector is expressed by placing
the vector symbol in absolute value brackets.
magnitudes of the vectors
The magnitude of a vector is a positive scalar.
u, v
Often it is necessary to explicitly state the initial
AB is the vector that starts
point and the end point of a vector. Then, two
at point A and ends at point B.
capital letters are used. Such vectors are referred
to as pointtopoint vectors.
/
/u
I u ,  v  are the
Its magnitude is ab.
Certain other terms are used in connection with vectors.
Two vectors are equal if and only if their magnitudes and their directions
are the same.
Two vectors are opposite if they have the same magnitude but point in
opposite directions.
When two vectors are opposite, such as /t/jf and CD,
one is the negative of the other: AB = —CD.
Two vectors are parallel if their directions are either the same or opposite.
EXAMPLE 2
ABCDEF is a regular hexagon. Give examples of vectors which are
a. equal
b. parallel but having different magnitudes
6
CHAPTER 1
A
f<
B
12.
c. equal in magnitude but opposite in direction
d. equal in magnitude but not parallel
e.
different in both magnitude and direction
Solution
a. AB = ~ED
b. TA~EB. but ~E # Ib
c. 7i = CB,butF£= CB
d. ~Ed = Z)c,but£D * ~DC
e.
FB,DC
There are other possible answers.
There is no special symbol for the direction of a vector. To specify the direction
of a vector, we state the angle it makes with another vector or with some given
direction such as a horizontal or vertical axis or a compass direction.
The angle between two vectors is the angle (< 180°) formed when the
vectors are placed tail to tail; that is, starting at the same point.
One way to determine the angle between two vectors is to examine geometrical
relationships and use trigonometry.
EXAMPLE 3
OABC is a square with sides measuring 6 units. E is the midpoint of BC.
Find the angle between the following vectors.
a. ~OB and OC
b. ~OE and ~OC
a
c. ~OB and ~OE
Solution
a. The diagonal of the square bisects ZAOC.
The angle between ~OB and OC is 45°.
b. Using trigonometry, tan ZEOC = r, ZEOC = 26.6°, so the angle between
~OE and ~OC is 26.6°.
c. The angle between ~OB and ~OE is the difference 45°  26.6° = 18.4°.
1.1 VECTOR CONCEPTS
13.
When two vectors are parallel, one of the vectors can be expressed in terms of the
other using scalarmultiplication. Suppose, for example, M is the midpoint of the
line segment Afl. Since M is the midpoint, then AB I = 21 AM , and since the
directions of AB and AM are the same, we write the vector equations
AB = 2AM or AM = jAB or ~BM = ~AB.
Thus, multiplication of a vector by a scalar k results in a new vector parallel to the
original one but with a different magnitude. It is true in general that two vectors
// and v are parallel if and only if it — kv.
A particularly useful type of vector is a vector with magnitude 1. Such vectors are
called unit vectors. A unit vector is denoted by a carat O placed over the symbol.
When a vector and a unit vector are denoted by the same letter, for example v and
v, you should understand v to be a unit vector having the same direction as v. Any
vector can be expressed as a scalar multiple of a unit vector.
Unit Vectors
1. A unit vector in the direction of any vector r can be found by
dividing v by its magnitude 11' :
i »■ I
...
2. Any vector v can be expressed as the product of its magnitude I v  and
a unit vector v in the direction of v:
i' =
r
r
Another useful type of vector has magnitude 0. Such vectors are valuable even
though their direction is undefined. The zero vector is denoted by 0.
EXAMPLE 4
Examine the vectors in the diagram.
a. Express b and c each as a scalar multiple of a.
b. Express «, b, and c each in terms of the unit vector a.
Solution
a. On the grid, each vector lies on the hypotenuse of a rightangled triangle
with sides in_the ratio 1:2, so the three vectors are parallel. The magnitudes
of a, b, and c can be found using the Pythagorean Theorem.
a = Vr + 22 = VS. T> = V52+ IO2 = 5V5,
and  c  = V32 + 62 = 3V5
Therefore b = 5a and c = 3a.
b. The unit vector in the direction of a is a = ~j=a. Then a =
8
CHAPTER 1
Tj = 5V5«, and c = V5
14.
Part A
1. In your own words, explain the difference between a scalar and a vector.
2. Which of these physical quantities is a vector and which is a scalar?
a. the acceleration of a drag racer
b. the mass of the moon
c. the velocity of a wave at a beach
d. the frequency of a musical note
e. the speed of light
f.
g. the friction on an ice surface
h. the volume of a box
i.
j.
the force of gravity
1.
the momentum of a curling stone
the energy produced by an electric
the age of a child
generator
k. the speedometer reading in an
automobile
m. the time on a kitchen clock
n. the magnetic field of the earth
o. the density of a lead weight
p. the pressure of the atmosphere
q. the area of a parallelogram
r.
the temperature of a swimming pool
3. For each part of Example 2, state a second answer.
Part B
4. One car travelling 75 km/h passes another going 50 km/h. Draw vectors that
represent the velocities of the two cars if they are going
a.
in the same direction
b.
in opposite directions
5. What is the angle between the following directions?
a. N and NE
b.
E and SW
c. 5 and W
6. Draw a vector to represent
a. the velocity of a fishing boat travelling at 8 knots on a heading of S 75° VV
(A knot is a speed of one nautical mile per hour.)
b.
the position of a city intersection 7 blocks east and 3 blocks south of your
present position
c.
the displacement of a crate that moves 6 m up a conveyor belt inclined at
an angle of 18°
d. the force exerted by a chain hoist carrying a load of 200 kg
1.1 VECTOR CONCEPTS
15.
7. Radar in the control tower of an airport shows aircraft at directions of N 50° E,
W70° W, and 5 20° W, and distances of 5, 8, and 12 km, respectively.
a. In a diagram, draw vectors showing the positon of the three aircraft
in relation to the tower.
b. The aircraft are travelling at velocities of 450 kph N, 550 kph N 70° W,
and 175 kph N 20° E, respectively. At the positon of each aircraft in part a,
draw small vectors to represent their velocities.
8. The points A, B, C, D, E, F, and G are equally spaced along a line. Name a
vector which is equal to
a. 3BD
b. jEA
c. yDF
d.
jGC
9. ABCD is a rhombus. For each of the following, find two
vectors it and v in this diagram (expressed as
e. 2AD
a
"
p
~
pointtopoint vectors) such that
a. it = v
b. it = —v
c. u = 2v
d.
it = jv
10. During takeoff, an aircraft rises 100 m for every 520 m of horizontal motion.
Determine the direction of its velocity.
11. Determine the magnitude and the direction of each of the
:::::;/
vectors in the given diagram. Express each direction as
an angle measured counterclockwise from a unit vector
in the positive x direction.
12. A search and rescue aircraft, travelling at a speed of 240 km/h, starts out at a
heading of N 20° W. After travelling for I h 15 min, it turns to a heading of
Af 80° E and continues for another 2 hours before returning to base.
a.
Determine the displacement vector for each leg of the trip.
b. Find the total distance the aircraft travelled and how long it took.
PartC
13. For what values of it is  [k  2)v < 14v , (v ¥= 0)?
14. Prove that two vectors u and v are parallel if and only if it = k!.
10
chapter i
16.
Section 1.2 —
In many applications of vectors to physical problems, we must find the combined
effect or sum of two or more vectors. What, for example, is the combined effect of
two or more forces acting on an object? How does wind velocity affect the
velocity of an aircraft?
To determine the sum of two vectors, let us look first for
a geometrical answer. Suppose the rectangle ABCD is
a park at the corner of an intersection. To get from A to C,
some people will walk along the sidewalk from A to B and
then from B to C. They follow a route described by the sum
e
of two displacement vectors: AB + BC. Others may follow
a shortcut through the park directly from A to C. This route
is described by the displacement vector AC.
Whichever route is followed, the displacement is the same;
both get from A to C. Therefore, AB + BC = AC.
vector
diagram
This model for vector addition is valid for all vectors because, in general, vectors
can be represented geometrically by a directed line segment.
TViangle Law of Vector Addition
To find the sum of two vectors u and v using the triangle law of vector
addition, draw the two vectors head to tail. The sum u + r. or resultant,
is the vector from the tail of the first to the head of the second.
The order in which we add the vectors
is unimportant. If the vectors are
added in the opposite order, the result
is the same. This demonstrates that
vectors satisfy the commutative law
of addition: u + v — v + it.
By combining the two triangles of the triangle law in one
diagram, a parallelogram is formed.
1.2 VECTOR LAWS
11
17.
Parallelogram Law of Vector Addition
To find the sum of two vectors using the parallelogram law of vector
addition, draw the two vectors tail to tail. Complete the parallelogram with
these vectors as sides. The sum it + v is the diagonal of the parallelogram
from the point where the tails are joined to the point where the heads meet.
These two laws of addition are equivalent. The method we use depends on which
is the most convenient for the problem at hand. When you set out to solve
a problem involving vectors, start by drawing vector diagrams such as those on
page 11.
EXAMPLE 1
Given the three vectors a, b, and c, sketch the sums a + b
and (a + Tj) + c b~ +c, a + (b + c).
Solution
Adding a to b first, we obtain
(a + b) + c
Adding b to c first, we obtain
b + c
a + (b + c)
This example illustrates that vectors satisfy the associative law of addition:
a + (b + c) = (a + b) + c. It means that we can omit the brackets and write
simply a + T) + c.
EXAMPLE 2
Find the magnitude and direction of the sum of two vectors u and v, if their
magnitudes are 5 and 8 units, respectively, and the angle between them is 30°.
Solution
Make a vector diagram showing the two vectors
with an angle of 30° between them. Complete the
parallelogram and draw the resultant.
12
chapter i
18.
The resultant is the third side of a triangle with sides 5
and 8. Observe that the angle between the vectors is
not an angle in this triangle. The angle between the
vectors is equal to an exterior angle of the triangle.
150°
8
(Why?) Use the angle of 150° and the cosine law to find the magnitude of the sum.
7i + rh = 52 + 82 2(5)(8)cos 150°
= 158.28
Then h + v = 12.6
The direction of u + v is expressed as an angle measured relative to one of the
given vectors, say v. This is 0 in the diagram. It can be found using the sine law.
sin 0 _ sin 150°
5
12.6
8= 11.4°
Therefore, the magnitude of u + v is 12.6 units, and it makes an angle of
approximately 11.4° with v.
To subtract two vectors a and /;, we express the difference in terms of a sum.
To find the vector a  b, use the opposite of b and add it to a. Hence a — bis
equivalent to a + (b).
The difference of two equal vectors a — a is the zero vector, denoted by 0.
The zero vector has zero magnitude. Its direction is indeterminate.
EXAMPLE 3
In parallelogram ABCD, find the difference AB — AD
a. geometrically
b.
algebraically
Solution
a.
Draw AD' opposite to AD. Using the
parallelogram law, draw the sum AB + AD
which is AC in the diagram.
But
= DB, so AB  AD = DB
b. AB  AD = AB + (AD)
= AB + DA
(DA ib the opposite; of AD)
= DA + AB
(rearrange the order of the terms)
= ~DB
1.2 VECTOR LAWS
13
19.
In the parallelogram formed by two vectors u and v
• the sum u + v is the vector created by the diagonal
from the tail of the two vectors
• the difference « — v is the vector created by the
second diagonal
u v =
Properties of Vector Addition
0a + b = b + a
Commutative Law
° (a +1>) + c = a + (b + c)
Associative Law
Properties of Scalar Multiplication
0 (mu)a = m(na)
Associative Law
® m(a + b) = ma + mb
Distributive Laws
• (m+n)a = ma + na
Properties of the Zero Vector: 0
aa + 0 = a
Each vector a has a negative (—a) such that
a + (a) = 0.
These laws state that you may add vectors in any order you like and that you may
expand and factor expressions in the usual way.
There are other basic vector relations that are universally true. We can
demonstrate the validity of these relations by using vector diagrams.
The following example illustrates this.
EXAMPLE 4
Show that m + v ^ u + v. When does equality hold?
Solution
Make a diagram of two vectors m and v, and their sum u + v.
The three vectors form a triangle. The lengths of the sides
of the triangle are the magnitudes of the vectors. From the
diagram, the side I u + v must be less than the sum of the
other two sides  u + v. There is no triangle if it is greater.
Therefore,
14
CHAPTER I
< 7i + v
20.
When u and v have the same direction, the triangle collapses to
a single line, and  u 4 v I =  //1 +  v .
TViangle Inequality
For vectors // and v, u + v I <  //1 4  v .
Part A
1. For each of the following, state the name of a vector equal to u + v and equal
to u — v.
2. Seven points A, B, C, D, E, F, and G, are arranged in order from left to right
on a single straight line. Express the vector BE as
a. the sum of two vectors, three vectors, and four vectors
b. the difference of two vectors in two different ways
3. What single vector is equivalent to each of these sums?
a. Jf+TS + 'SQ
b. A~CG~E + ~CE
c. EA~CB + ~DB + ~AD
d. PT  gf + 51?  Sg
PartB
4. Find the sum of the vectors w and v if 6 is the angle between them.
a. h = 12, v = 21, and 0 = 70° b. 17t = 3,  v = 10, and 9 = 115°
5. A tour boat travels 25 km due east and then 15 km 5 50° E. Represent these
displacements in a vector diagram, then calculate the resultant displacement.
6. If a and b are unit vectors that make an angle of 60° with each other, calculate
a. 3«5£
b. 8« + 36
1.2 VECTOR LAWS
15
21.
7. What conditions must be satisfied by the vectors 7i and v for the following to
be true?
a.  u + v  =  u  v 
b.  u + v  >  u  v 
c.  u + v < u  v
8. Under what conditions will three vectors having magnitudes of 7, 24, and 25,
respectively, have the zero vector as a resultant?
9. Vectors a and b have magnitudes 2 and 3, respectively. If the angle between
them is 50°, find the vector 5a  2b, and state its magnitude and direction.
10. Simplify the following expressions using the properties of vector operations,
a. 4x5yx + 6y
b. 2x  4(.v  y)
c. 8(3^ + 5y)  4(dv  9v)
d. 3.v  6v + 4(2y  .v)  6.v
11. Let a = 2/  3J + k, b = 1 + j + it, and c = 2/  3k. Find
a. a + b + c
b. a + 2b  3c
c.
3b + Ac
12. If a = 3.t + 2) and b = 5.v  4j, find x and y in terms of a and b.
13. Check each identity algebraically, and illustrate with the use of a diagram.
a.
y  .r _ .v + v
v + ■ 2
2 •
,
b.
x
x + v _ .v  y
—y~  ~y~
14. Illustrate for k > 0 that k(u + v) = k~i + kv.
15. Show geometrically that, for any scalar k and any vectors u and v,
it(i/  ') = ku — At.
16. By considering the angles between the vectors, show that a + T> and a  b are
perpendicular when a = b.
PartC
17. ABCDEF is a regular hexagon with sides of unit length.
Find the magnitude and the direction of
18. If jc = 11, y =23, and xy = 30, find .v + y.
19. The sum and the difference of two vectors u and v are given.
Show how to find the vectors themselves.
16
CHAPTER 1
U + V
I UV
22.
20. Represent by i,j, and k the three vectors AB, AC,
and AD that lie along adjacent edges of the cube in the
given diagram. Express each of the following vectors in
r
terms of?,/, and k.
a.
'
A>
*
FG, a diagonal of the front face of the cube
h
7
b. the other diagonals of the front, top, and right faces of the cube
c. BE, a body diagonal of the cube
d. the other body diagonals of the cube
c. What is the magnitude of a face diagonal? A body diagonal?
21. Prove that for any vectors 7i and v, 17t + v 2 +  w  v 2 = 2( 7t 2 +  v 2).
1.2 VECTOR LAWS
17
23.
Section 13 —
A force on any object causes that object to undergo an acceleration. You can feel
a force pushing you back into your seat whenever the car you are riding in
accelerates from a stop light. You no longer feel any force once the car has
reached a steady speed, but that does not mean that the force that set the car
in motion has ceased to exist. Instead that force is now balanced by other forces
such as air resistance and road friction. A steady speed is an example of a
state of equilibrium in which the net force is zero.
It was Newton who first clarified these concepts and formulated the law that bears
his name.
Newton's First Law of Motion
An object will remain in a state of equilibrium (which is a state of rest
or a state of uniform motion) unless it is compelled to change that state
bv the action of an outside force.
The outside force mentioned in Newton's First Law refers to an unbalanced force.
When you release a heliumfilled balloon, it will rise into the air. It is attracted by
the force of gravity like everything else but upward forces are greater, so it
accelerates into the sky. Eventually it reaches an altitude where the atmosphere is
less dense, and the buoyant forces and the force of gravity balance. In this state
of equilibrium, it can float for days, as weather balloons often do.
EXAMPLE 1
Describe the forces acting on an aircraft flying at constant velocity.
Solution
An aircraft flying at a constant velocity is in a state of
equilibrium. The engines provide thrust, the force
propelling the aircraft forward. The thrust is counter
aft
f
drag •«— aircraft —*■ thrust
balanced by a drag force coming from air resistance.
J,
The air rushing past the wings produces lift, a force
weight
which counterbalances the force of gravity and keeps
the plane aloft.
The magnitude of a force is measured in newtons, which is abbreviated as N. At
the earth's surface, gravity causes objects to accelerate at a rate of approximately
9.8 m/s2 as they fall. The magnitude of the gravitational force is the product of an
18
chapter i
24.
object's mass and this acceleration. The gravitational force on a 1kg object at the
earth's surface is approximately 9.8 N. In other words, a 1kg object weighs
approximately 9.8 N.
It is generally the case that several forces act on an object at once. It is important
to know the net effect of all these forces because an object's state of motion is
determined by this net force. Since forces are vectors, the single force that has the
same effect as all the forces acting together can be found by vector addition. This
single force is the resultant of all the forces.
Sometimes a force acts on an object at an angle, so that only part of the force is
affecting the motion of the object.
EXAMPLE 2
Jake is towing his friend on a toboggan, using a rope which makes an angle of 25°
with the ground. If Jake is pulling with a force of 70 N, what horizontal force is
he exerting on the toboggan?
Solution
First draw a diagram showing the force and its
direction. Now consider that this force is the
resultant of a horizontal force h and a vertical
force v. We show this by forming a triangle,
with the original 70 N force as the resultant;
h and v are perpendicular.
Now  Ti  = 70 cos 25°
So the horizontal force is about 63.4 N.
We refer to the quantieties  Tt  and v as the horizontal and vertical components
of the original force.
EXAMPLE 3
Jake and Maria are towing their friends on a toboggan. Each is exerting a
horizontal force of 60 N. Since they are walking side by side, the ropes pull one
to each side; they each make an angle of 20° with the line of motion. Find the
force pulling the toboggan forward.
Solution
Make a diagram showing the forces. By completing
the parallelogram, we show the resultant r.
the diagonal of the parallelogram.
1.3 FORCE AS A VECTOR
19
25.
r2 = 602 + 602  2(60)(60) cos 140°
7 = 112.8
The towing force is about 113 N.
1. We could have solved this question by finding the component of each
force along the direction of travel and adding the results.
2. If the forces had not been equal, the angles made with the direction
of travel would not have been equal.
In Example 3, the toboggan is (probably) travelling at a constant speed, indicating
that there is no unbalanced force on it. This is because there is a frictional force
that is equal and opposite to the towing force.
The force that is equal in magnitude but opposite in direction to the resultant is
called the equilibrant. It exactly counterbalances the resultant. In Example 2, the
force of friction is the equilibrant, which keeps the towing force from accelerating
the toboggan.
EXAMPLE 4
In Example 2, what if Maria starts pulling at an
angle of 30° instead of 20°? As the diagram
shows, the direction of the resultant will be a
little to the right of the axis of the toboggan. This
means that the toboggan will not travel forward in
Maria
a straight line but will veer continually to the right.
If these conditions remain unchanged, the toboggan
will travel in a circle.
EXAMPLE 5
Jake
In Example 2, if Maria pulls with
60
a force of 60 N at an angle of
30°, what should the magnitude  {
of the force exerted by Jake at
an angle of 20° be if the toboggan
is to move straight forward without
turning? According to the sine law,
sin 30° _ sin 20"
F
~
60
~F = 88N
20
CHAPTER 1
Maria
26.
Jake must pull with a force of 88 N. Since Jake is pulling harder than before, the
resultant will be greater than before:
sin 130"_ sin 20°
R
60
R= 134 N
As in Example 2 and the subsequent discussion, make it a practice with force
problems to look for ways to justify your numerical results and make them
physically meaningful.
EXAMPLE 6
A large promotional balloon is tethered to the top of a building by two guy wires
attached at points 20 m apart. If the buoyant force on the balloon is 850 N, and
the two guy wires make angles of 58° and 66° with the horizontal, find the tension
in each of the wires.
Solution
First draw the position diagram showing where the forces act. In this problem, the
resultant of the two tensions must be 850 N to counterbalance the buoyant force
of the balloon, which is the equilibrant. In making the force diagram, draw the
tension vectors parallel to the corresponding lines in the position diagram.
In the diagrams, observe step by step how the angles in the position diagram are
first translated into the force diagram, and then how these angles are used to
determine the angles inside the force triangle.
■24
850 N
position diagram
Since all three angles in the force triangle are known, the magnitudes of the
tension vectors 7", and T2 can be calculated using the sine law,
ft
sin 24°
Therefore,
_
850
sin 124°
_ 850 sin 24"
sin 124"
= 4I7N
A
and
and
T2
_
850
sin 124°'
sin 32°
_ 850 sin 32°
sin 124°
= 543 N
The tensions in the guy wires are approximately 417 N and 543 N, with the guy
wire at the steeper angle having the greater tension.
1.3 FORCE AS A VECTOR
21
27.
EXAMPLE 7
Is it possible for an object to be in a state of equilibrium when forces of 10 N,
20 N, and 40 N act on it?
Solution
An object will be in a state of equilibrium when the resultant of all the forces
acting on it is zero. This means that the three given force vectors must form a
triangle. By the triangle inequality theorem, the sum of any two sides must be
greater than the third, but in this case the magnitudes of the forces are such that
10 + 20 < 40. Therefore, an object cannot be in a state of equilibrium with the
three given forces acting on it.
In the discussion of forces in the previous examples,
we assumed that an object is free to move in the
direction of the force acting on it. Often, however, that
is not the case. For example, when you push a lawn
mower, you exert a force along the handle, but the
mower does not move into the ground along the line of
the force. It moves horizontally. So, how much of the
motion
force that you exert actually contributes to the motion?
To answer this question, we must resolve the force into
n
horizontal and vertical components. The components are
. ,
vertical
component
the magnitudes of forces acting horizontally and vertically,
whose sum, by vector addition, is the original force.
horizontal
component
EXAMPLE 8
A lawn mower is pushed with a force of 90 N directed along the handle, which
makes an angle of 36° with the ground.
a. Determine the horizontal and vertical components of the force on the
mower.
b. Describe the physical consequences of each component of the pushing
force.
Solution
a. The force diagram is a right triangle.
The components are
Fh = 90 cos 36°
= 72.8 N
and
FV = 90 sin 36°
= 52.9 N
b. The horizontal component of the force, 72.8 N, moves the lawnmower
forward across the grass. The vertical component of the force, 52.9 N, is in
the same direction (down) as the force of gravity.
22
chapter i
28.
EXAMPLE 9
A 20kg trunk is resting on a ramp inclined at an angle of 15°. Calculate the
components of the force of gravity on the trunk that are parallel and perpendicular
to the ramp. Describe the physical consequences of each.
Solution
The force of gravity on the trunk is (20 kg) x (9.8 m/s2) = 196 N acting down.
The parallel and perpendicular components are
I?, = 196 sin 15° and
= 51N
fJ = 196 cos 15°
= 189N
_
fi96N
position diagram
196 N
force diagram
The parallel component points down the slope of the ramp. It tends to cause the
trunk to slide down the slope. It is opposed by the force of friction acting up the
slope. The perpendicular component presses the trunk against the ramp. The
magnitude of the force of friction is proportional to this component.
Part A
1. Name some common household objects on which the force of gravity is
approximately 2 N; 20 N; 200 N. What is your weight in newtons?
2. Find the horizontal and vertical components of each of the following forces.
a. 200 N acting at an angle of 30° to the horizontal
b.
160 N acting at an angle of 71° to the horizontal
c. 75 N acting at an angle of 51° to the vertical
d. 36 N acting vertically
3. Find the resultant of each pair of forces acting on an object.
a.
forces of 7 N east and 12 N west
b. forces of 7 N east and 12 N north
c. forces of 6 N southwest and 8 N northwest
d. forces of 6 N southeast and 8 N northwest
1.3 FORCE AS A VECTOR
23
29.
PartB
9N
4. Find the magnitude of the resultant of the four forces
10 n
7N
shown in the given diagram.
5. Two forces /•"] and F2 act at right angles to each other. Express the magnitude
and direction of F, + F2 in terms of  /r11 and  F2 I.
6. Find the magnitude and the direction (to the nearest degree) of the resultant
of each of the following systems of forces.
a. forces of 3 N and 8 N acting at an angle of 60° to each other
b. forces of 15 N and 8 N acting at an angle of 130° to each other
7. Find the magnitude and direction of the equilibrant of each of the following
systems of forces.
a.
forces of 32 N and 48 N acting at an angle of 90° to each other
b. forces of 16 N and 10 N acting at an angle of 10° to each other
8. Is it easier to pull yourself up doing chinups when your hands are 60 cm
apart or 120 cm apart? Explain your answer.
9. A mass of 10 kg is suspended from a ceiling by two cords that make angles
of 30° and 45° with the ceiling. Find the tension in each of the cords.
10. Two forces of equal magnitude act at 60° to each other. If their resultant has
a magnitude of 30 N, find the magnitude of the equal forces.
11. Which of the following sets of forces acting on an object could produce
equilibrium?
a. 5N, 2N, 13 N
c.
13 N, 27 N, 14 N
b. 7N, 5N, 5N
d.
12 N, 26 N, 13 N
12. Three forces of 5 N, 7 N, and 8 N are applied to an object. If the object is in
a state of equilibrium
a.
show how the forces must be arranged
b. calculate the angle between the lines of action of the 5 N and 7 N forces
13. A man weighing 70 kg lies in a hammock whose ropes make angles of
20° and 25° with the horizontal. What is the tension in each rope?
14. A steel wire 40 m long is suspended between two fixed points 20 m apart.
A force of 375 N pulls the wire down at a point 15 m from one end of the
wire. State the tension in each part of the wire.
24
chapter 1
30.
15. An advertising sign is supported by a horizontal steel brace
extending at right angles from the side of a building, and by
25°
a wire attached to the building above the brace at an angle
of 25°. If the force of gravity on the sign is 850 N, find the
u
tension in the wire and the compression in the steel brace.
16. Find_the .v and ycomponents of each of the vectors », v,
and u*.
17. A tractor is towing a log using a cable inclined at an angle of 15° to the
horizontal. If the tension in the cable is 1470 N, what is the horizontal force
moving the log?
18. A piece of luggage is on a conveyer belt that is inclined at an angle of 28°.
If the luggage has a mass of 20 kg
a. determine the components of the force of gravity parallel to and
perpendicular to the conveyer belt
b. explain the physical effect of each of these components
19. A child with a mass of 35 kg is sitting on a swing attached to a tree branch by
a rope 5 m in length. The child is pulled back 1.5 m measured horizontally.
a. What horizontal force will hold the child in this position?
b. What is the tension in the rope?
20. The main rotor of a helicopter produces a force of 55 kN. If the helicopter
flies with the rotor revolving about an axis tilted at an angle of 8° to the
vertical
a. find the components of the rotor force parallel to and perpendicular
to the ground
b. explain the physical effect on the helicopter of each component of the
rotor force
21. In order to keep a 250kg crate from sliding down a ramp inclined at 25°, the
force of friction that acts parallel to and up the ramp must have a magnitude
of at least how many newtons?
22. A lawn roller with a mass of 50 kg is being pulled with a force of 320 N.
If the handle of the roller makes an angle of 42° with the ground, what
horizontal component of the force is causing the roller to move?
1.3 FORCE AS A VECTOR
25
31.
PartC
23. Three forces, each of which is perpendicular to the
other two, act on an object. If the magnitudes of
ION
these forces are 6 N, 15 N, and 10 N, respectively,
find the magnitude and direction of the resultant.
(State the angles that the resultant makes with the
15N
6N,
two larger forces.)
24. Two tugs are towing a ship. The smaller tug is 10° off the port bow and the
larger tug is 20° off the starboard bow. The larger tug pulls twice as hard as
the smaller tug. In what direction will the ship move?
25. Braided cotton string will break when the tension exceeds 300 N. Suppose
that a weight of 400 N is suspended from a 200cm length of string, the upper
ends of which are tied to a horizontal rod at points 120 cm apart.
a.
Show that the string will support the
•120cm
weight, when the weight is hung at the
centre of the string.
100 cm
100 cm
400 N
b. Will the string break if the weight is 80 cm from one end of the string?
26
chapter i
32.
In elementary problems, the speed of a moving object is calculated by dividing
the distance travelled by the travel time. In advanced work, speed is defined more
carefully as the rate of change of distance with time. In any case, speed is a
quantity having magnitude only, so it is classified as a scalar.
When the direction of motion as well as its magnitude is important, the correct
term to use is velocity. Velocity is a vector quantity. Speed is the magnitude
of a velocity.
Velocity vectors can be added. When you walk forward in the aisle of an aircraft
in flight, the 2km/hr velocity of your walk adds to the 500km/hr velocity of the
plane, making your total velocity 502 km/hr. When two velocities are not in the
same direction, the resultant velocity determined from the addition of two velocity
vectors is nevertheless a meaningful, physical quantity.
EXAMPLE 1
A canoeist who can paddle at a speed of 5 km/h in still water
wishes to cross a river 400 m wide that has a current of 2 km/h.
If he steers the canoe in a direction perpendicular to the current,
determine the resultant velocity. Find the point on the opposite
bank where the canoe touches.
Solution
As the canoe moves through the water, it is carried
vector diagram
sideways by the current. So even though its heading is
5 km/h
straight across the current, its actual direction of motion
2 km/h
is along a line angling downstream determined by the
sum of the velocity vectors.
From the vector diagram,
 r 2 = (5)2 + (2)2
and
tan 6 = j
I v = V29 = 5.4 km/h
8 = 21.8°
Therefore, the canoeist crosses the river at a speed of 5.4 km/h along a line at an
angle of about 22°. The displacement triangle is similar to the vector triangle.
400 m
5 km/h
2 km/h
x_ _ 400
2
5
.v= 160
1.4 VELOCITY AS A VECTOR
27
33.
He touches the opposite bank at a point 160 m downstream from the point directly
opposite his starting point. We could also find .v using the angle 6, but we must be
careful not to round off in the process.
EXAMPLE 2
Suppose the canoeist of Example 1 had wished to travel straight
across the river. Determine the direction he must head and the
time it will take him to cross the river.
Solution
In order to travel directly across the river, the canoeist must
vector diagram
steer the canoe slightly upstream. This time, it is the vector
sum, not the heading of the canoe, which is perpendicular
2km/h
to the river bank. From the vector diagram,
 (2)2
and
= 4.6 km/h
sin (0) = =■
6 = 23.6°
Therefore, to travel straight across the river, the canoeist must head upstream at an
angle of about 24°. His crossing speed will be about 4.6 km/h.
The time it takes to cross the river is calculated from
_
river width
crossing speed
 JM_
(where the width is 0.4 km)
(we avoid using rounded values if possible)
= 0.087 h or 5.2 min
It takes the canoeist approximately 5.2 minutes to cross the river.
Wind affects a plane's speed and direction much the same way that current affects
a boat's. The airspeed of a plane is the plane's speed relative to the mass of air it
is flying in. This may be different in both magnitude and direction from the
plane's ground speed, depending on the strength and direction of the wind.
EXAMPLE 3
An airplane heading northwest at 500 km/h encounters a wind of 120 km/h from
25° north of east. Determine the resultant ground velocity of the plane.
Solution
Since the wind is blowing from 25° north of east, it can be represented by a vector
whose direction is west 25° south. This wind will blow the plane off its course.
28
chapter i
34.
changing both its ground speed and its heading. Let  v  be the airspeed of the
plane and  w be the wind speed. On a set of directional axes, draw the two
velocity vectors. Then draw the resultant velocity using the parallelogram law
of vector addition.
plane heading
W
wind direction
In parallelogram OCBA, ZCOA = 45° + 25° = 70°, so ZOAB = 110°. Then, in
AOAB, two sides and the included angle are known, so the magnitude of the
resultant velocity can be calculated using the cosine law.
I v + uM2 = 5002 + 1202  2(5OO)(12O) cos 110°
 v + vr  s 552.7
Store this answer in your calculator memory.
500
Next, ZAOB can be calculated from the sine law.
sin ZAOB _
500
sin 110°
v + ^
JSt
the value of
i_a!a.!ated above)
ZAOB = 58.2°
ZWOB = 58.2°  25° = 33.2°
The resultant velocity has direction 33° north of west and a magnitude
of553km/h.
A key step in solving problems such as that in Example 3 is to find an angle
in the triangle formed by the vectors. Here is a helpful hint: identify which
angle is formed by vectors whose directions are given, and draw small axes at
the vertex of that angle. The diagram shows this alternate way to calculate that
ZOAB = 110° in Example 3.
Vectors are needed to describe situations where two objects are moving relative to
one another. When astronauts want to dock the space shuttle with the International
Space Station, they must match the velocities of the two craft. As they approach,
astronauts on each spacecraft can picture themselves to be stationary and the other
craft to be moving. When they finally dock, even though the two spacecraft are
orbiting the earth at thousands of miles per hour, their relative velocity is zero.
1.4 VELOCITY AS A VECTOR
29
35.
Relative velocity is the difference of two velocities. It is what an observer
measures, when he perceives himself to be stationary. The principle that all
velocities are relative was originally formulated by Einstein and became a
cornerstone of his Theory of Relativity.
When two objects A and B have velocities v/t and vw, respectively,
the velocity of B relative to A is
vnl = vtt ~ v,
EXAMPLE 4
A car travelling east at 110 km/h passes a truck going in the opposite direction
at 96 km/h.
a. What is the velocity of the truck relative to the car?
b. The truck turns onto a side road and heads northwest at the same speed.
Now what is the velocity of the truck relative to the car?
Solution
The vector diagram shows the velocity vectors of the car and
the truck. These velocities are relative to someone standing by
*~^
truck
the side of the road, watching the two vehicles pass by. Since
1
lr"ck
car
v
the car is going east, let its velocity be vC(lr = 110. Then the
truck's velocity is vlruck = 96.
'—j—
rel
v
truck
v
car
= (96)(110)
= 206 km/h or 206 km/h west
This is the velocity that the truck appears to have, according to the driver of the car.
b. After the truck turns, the angle between the car and the
truck velocities is 135°. The magnitude of the sum is
found using the cosine law.
vn., 2 = (96)2 + (110)2 — 2(96)(110) cos 135°
vw/ = 190.4 km/h
(Store this in your calculator.)
The angle of the relative velocity vector can be calculated from the sine law.
sin 9 _ sin 135°
96
190.4
8 = 20.9°
30
CHAPTER 1
v
car
36.
After the truck turns, its velocity is 190 km/h in a direction W2° N relative to the
car. Note that the relative velocity of the two vehicles does not depend on their
position. It remains the same as long as the two vehicles continue to travel in the
same directions without any changes in their velocities.
Part A
1. A plane is heading due east. Will its ground speed be greater than or less than
its airspeed, and will its flight path be north or south of east when the wind
is from
a.
N
b. SS0°W
c.
S30°£
d.
N 80° E
2. A man can swim 2 km/h in still water. Find at what angle to the bank he must
head if he wishes to swim directly across a river flowing at a speed of
a.
1 km/h
b. 4 km/h
3. A streetcar, a bus, and a taxi are travelling along a city street at speeds of 35,
42, and 50 km/h, respectively. The streetcar and the taxi are travelling north;
the bus is travelling south. Find
a. the velocity of the streetcar relative to the taxi
b. the velocity of the streetcar relative to the bus
c.
the velocity of the taxi relative to the bus
d. the velocity of the bus relative to the streetcar
PartB
4. A river is 2 km wide and flows at 6 km/h. A motor boat that has a speed of
20 km/h in still water heads out from one bank perpendicular to the current.
A marina lies directly across the river on the opposite bank.
a. How far downstream from the marina will the boat reach the other bank?
b.
How long will it take?
5. An airplane is headed north with a constant velocity of 450 km/h. The plane
encounters a west wind blowing at 100 km/h.
a. How far will the plane travel in 3 h?
b. What is the direction of the plane?
6. A light plane is travelling at 175 km/h on a heading of N 8° E in a 40km/h
wind from /V 80° E. Determine the plane's ground velocity.
1.4 VELOCITY AS A VECTOR
31
37.
7. A boat heads 15° west of north with a water speed of 3 m/s. Determine its
velocity relative to the ground when there is a 2 m/s current from 40° east
of north.
8. A plane is steering east at a speed of 240 km/h. What is the ground speed of
the plane if the wind is from the northwest at 65 km/h? What is the plane's
actual direction?
9. Upon reaching a speed of 215 km/h on the runway, a jet raises its nose to an
angle of 18° with the horizontal and begins to lift off the ground.
a. Calculate the horizontal and vertical components of its velocity
at this moment.
b. What is the physical interpretation of each of these components
of the jet's velocity?
10. A pilot wishes to fly to an airfield 5 20° E of his present position. If the
average airspeed of the plane is 520 km/h and the wind is from N 80° E
at 46 km/h,
a.
in what direction should the pilot steer?
b. what will the plane's ground speed be?
11. A destroyer detects a submarine 8 nautical miles due east travelling northeast
at 20 knots. If the destroyer has a top speed of 30 knots, at what heading
should it travel to intercept the submarine?
PartC
12. An airplane flies from Toronto to Vancouver and back. Determine which time
is shorter.
a. The time for the round trip when there is a constant wind blowing from
Vancouver to Toronto
b. The time for the round trip when there is no wind
13. A sailor climbs a mast at 0.5 m/s on a ship travelling north at 12 m/s, while
the current flows east at 3 m/s. What is the speed of the sailor relative to the
ocean floor?
14. A car is 260 m north and a truck is 170 m west of an intersection. They are
both approaching the intersection, the car from the north at 80 km/h, and the
truck from the west at 50 km/h. Determine the velocity of the truck relative to
the car.
32
chapter i
38.
In this chapter, you have been introduced to the concept of a vector and have seen
some applications of vectors. Perhaps the most important mathematical skill to
develop from this chapter is that of combining vectors through vector addition,
both graphically and algebraically.
Diagrams drawn free hand are sufficient, but try to make them realistic. It is not
difficult to draw angles that are correct to within about 10° and to make lengths
roughly proportional to the magnitudes of the vectors in a problem.
Once you have calculated answers, ask yourself if the calculated angles and
magnitudes are consistent with your diagram, and if they are physically
reasonable.
SUMS
Speaking informally, if you want to go from A to C you
can travel directly along the vector AC, or you can detour
through B, travelling first along AB, and then along BC.
This means that AC = AB + BC, but observe how the
detour point fits into the equation: it is the second letter
of the first vector and the first letter of the second vector.
DIFFERENCES
Using the same diagram, if you want to go from D to B, you can travel directly
along DB, or you can detour through A, travelling first backwards along A~D, and
then forwards along AB. This translates into the equation DB = AD + AB,
which of course is just the difference DB = AB  AD. Note carefully that, on the
right hand side of the equation, the order of the initial point D and the end point B
are reversed, and the detour point is the initial letter of the two vectors.
Pay attention to and become familiar with details such as these. You will be able
to draw and interpret vector diagrams and handle vector equations more quickly
and correctly if you do.
KEY CONCEPTS REVIEW
33
39.
CHAPTER 1: VECTORS AND^flE SUPERIOR COLLICULUS
Brain cells in the superior colliculus are tuned to the directions of distant visual
and auditory stimuli. Each cell responds only to stimuli located within a cone of
directions. The vigour of a cell's response can be regarded as specifying the
magnitude of a vector in the direction the cell represents. The resultant vector
formed by summing the vectors represented by the individual cells points in the
direction of the stimulus.
Dr. Randy Gailistel, a professor in the Department of Psychology at UCLA, whose
research focus is in the cognitive neurosciences, has suggested that these
neurological resultant vectors are "the first new idea about how the nervous
system represents the value of a variable since the beginning of the [twentieth]
century (from Conservations in the Cognitive Neurosciences, Ed. Michael S.
Gazzaniga, MA: Bradford Books/MIT Press, 1997)."
Investigate and Apply
1. What direction would be represented by a north cell responding three times
as vigorously as a northeast cell, which, in turn, is responding twice as
vigorously as an east cell?
2. Consider an ensemble of 36 cells, representing directions evenly distributed
around a circle, with one cell representing north. One cell will represent 10°
east of north, the next will represent 20° east of north, and so on. A cell
always responds to some extent whenever a stimulus is within 20° of the cell's
direction.
a) Which cells will respond to a stimulus whose direction is northeast?
b) A response pattern is a description of the relative proportions of the vigour
of the various cells' responses. Give two possible response patterns for the
cells found in part a.
3. How do you think the brain deals with the fact that several different response
patterns can represent the same direction?
INDEPENDENT STUDY
Investigate the field of neuroscience.
What other things can be represented in the brain using resultant vectors formed
from cells representing individual vectors?
What are some other questions to which neuroscientists are seeking answers?
What role does mathematics play in the search for answers to these questions? ®
V
34
CHAPTER 1
40.
Reiriei
1. a. If v + t = v. what is /?
b.
If tv = v, what is t'l
c. If sv = tit, and 1/ is not parallel to v, what are s and /?
2. Using vector diagrams, show that
a. (a + b)u = an + bit
b. (ab)7i = a(b7t)
3. A mass M is hung on a line between two supports A and B.
a.
A
Which part of the line supporting the mass has the
greater tension? Explain.
b. The supports A and B are not at the same level. What
effect does this have on the tension in the line? Explain.
4. Explain these properties of the zero vector:
a. Ov = 0
b. v + 0 = v
c. if 7i + v = 0, then m = v
5. If/ andy are perpendicular unit vectors, what is the magnitude of
a. 3/ + 4/?
b. 24/  7/?
c. ai + bp.
6. Show that a + b~ =  a — b , if a and b have opposite directions.
7. A 3kg mass is hanging from the end of a string. If a horizontal force of 12 N
pulls the mass to the side
a. find the tension in the string
b. find the angle the string makes with the vertical
8. Two forces Fx and F2 act on an object. Determine the magnitude of the
resultant if
a. I F I = 54 /V, I F2 I = 34 N, and the angle between them is 55°
b. I Fx I = 21 N, I F21 = 45 N. and the angle between them is 140°
9. Two forces at an angle of 130° to each other act on an object. Determine their
magnitudes if the resultant has a magnitude of 480 N and makes an angle of
55° with one of the forces.
REVIEW EXERCISE
35
41.
10. Forces of 5 N, 2 N, and 12 N, all lying in the same plane, act on an object.
The 5 N and 2 N forces lie on opposite sides of the 12 N force at angles of
40° and 20°, respectively. Find the magnitude and direction of the resultant.
11. A 10kg mass is supported by two strings of length 5 m and 7 m attached
to two points in the ceiling 10 m apart. Find the tension in each string.
12. The pilot of an airplane that flies at 800 km/h wishes to travel to a city
800 km due east. There is a 80 km/h wind from the northeast.
a. What should the plane's heading be?
b. How long will the trip take?
13. An airplane heads due south with an air speed of 480 km/h. Measurements
made from the ground indicate that the plane's ground speed is 528 km/h at
15° east of south. Calculate the wind speed.
14. A camp counsellor leaves a dock paddling a canoe at 3 m/s. She heads
downstream at 30° to the current, which is flowing at 4 m/s.
a. How far downstream does she travel in 10 s?
b. What is the length of time required to cross the river if its width is 150 m?
15. A pilot wishes to reach an airport 350 km from his present position at
a heading of N 60° E. If the wind is from 5 25° E with a speed of 73 km/h.
and the plane has an airspeed of 450 km/h, find
a. what heading the pilot should steer
b. what the ground speed of the plane will be
c. how many minutes it will take for the plane to reach its destination
16. A coast guard cutter is steering west at 12 knots, when its radar detects a
tanker ahead at a distance of 9 nautical miles travelling with a relative
velocity of 19 knots, on a heading of £ 14° N. What is the actual velocity of
the tanker?
17. Twice a week, a cruise ship carries vacationers from Miami, Florida to
Freeport in the Bahamas, and then on to Nassau before returning to Miami.
The distance from Miami to Freeport is 173 km on a heading of E 20° N.
The distance from Freeport to Nassau is 217 km on a heading of E 50° S.
Once a week the ship travels directly from Miami to Nassau. Determine the
displacement vector from Miami to Nassau.
18. If an + bv = 0 and u and v have different directions, what must a and b
equal?
19. Show geometrically that  i<  v  ^ // + v . Under what conditions
does equality hold?
36
chapter 1
42.
1. Under what conditions is u + vI = u +  v ?
2. Copy the three given vectors a, b, and c onto
graph paper, then accurately draw the following
three vectors.
a.
b.
c.
u = a + 3c
v = b  a
—
■}—
—
—
w = ^b  5c + a
a
3. Simplify 3(4» + v)  2u  3(h  v).
4. Illustrate in a diagram the vector property 4(a + b) = Aa + 4b. What is this
property called?
5. Forces of 15 N and II N act on a point at 125° to each other. Find the
magnitude of the resultant.
6. A steel cable 14 m long is suspended between two fixed points 10 m apart
horizontally. The cable supports a mass of 50 kg at a point 6 m from one end.
Determine the tension in each part of the cable.
7. A ferry boat crosses a river and arrives at a point on the opposite bank
directly across from its starting point. The boat can travel at 4 m/s and the
current is 1.5 m/s. If the river is 650 rn wide at the crossing point, in what
direction must the boat steer and how long will it take to cross?
8. What is the relative velocity of an airplane travelling at a speed of 735 knots
on a heading of E 70° S with respect to an aircraft at the same height steering
W 50° 5 al a speed of 300 knots?
CHAPTER 1 TEST
37
43.
IEUCLIDEAW^EOMETRY
The word geometry comes from the Greek words for earth and measure. When we solve geometrical
problems, the rules or assumptions we make are chosen to match our experience with the world we live
in. For example, since locally the earth looks flat, it makes sense to talk about planar figures such as
triangles, circles, and so on. But what happens if we change the rules? For example, we normally define
distance in Euclidean terms. When we represent points and figures in terms of coordinates on the
Cartesian plane, then the distance between two points P(xx, y,) and Q(*2, y2) is
, Q) =
x2Y + (y,  y2)2
If we ask for the locus of all points that are a constant distance, say 1, from the given point (0,0), we
get the circle with equation x2 + y2 = 1.
One way to create a whole new geometry is to change the way we measure distance. For example, we
can use the socalled taxicab distance given by
t(P,Q)= U,jc2 + y,y2
Y2y
38
CHAPTER 1
44.
The taxicab distance between P and Q is the sum of the lengths PR and RQ. The reason for the
colourful name is that it is the actual distance driven if a cab is restricted to a rectangular grid of streets.
Note that t(P, Q) £ d(p, Q) for any pair of points P and Q.
With this definition of distance, we can ask the same locus question. What is the set of all points a taxicab distance of I from the origin? If P(x, y) is any point on the locus, then the equation of the locus is
x — 01 + y — 01 = 1 or I jc  + y =1. The locus is plotted below, and turns out to be a square.
The graph can be produced by a graphing calculator or by hand. In this case, it is easiest to break the
problem into four cases depending on x and y being positive or negative.
You can investigate many other locus problems in this new geometry. For example, find the set of
points that are equidistant from (0, 0) and (1, 1). If we use Euclidean distance, we get a straight line,
the right bisector of the line segment joining the two points. The following diagram shows what
happens with taxicab distance.
EXTENDING AND INVESTIGATING
39
45.
(1.1)
(0,0)
For Osxs ], the right bisector is the line, as with Euclidean distance. However, for.v s 1, y < 0 and
x ^ 0, y s 1, all points are equidistant from (0, 0) and (1, 1).
There are many other ways to generate nonEuclidean geometries. Another example is to look at
geometry on the surface of a sphere. In this geometry, straight lines (the shortest path between two
points) become arcs of circles.
For fun, try the following with taxicab distance:
1. Find an equilateral triangle with taxicab side length 1. Are all angles equal?
2. Sketch the locus of all points that are equidistant from (0, 0) and (1,2).
3. The line segment joining (0, 0) to (1, 0) is rotated about the origin. What happens to its length?
40
chapter 1
47.
CHAPTER 2: MOLECULAR B&ND ANGLES
Atoms bond together to form the molecules that make up the substances around
us. The geometry of molecules is a factor in determining many of the chemical
properties of these substances. Ethyl alcohol and dimethyl ether are both formed
from two carbon atoms, six hydrogen atoms, and one oxygen atom (C2H6O), but
they have very different chemical and physical attributes. The properties of
enzymes, protein molecules that speed up biochemical reactions, depend upon
precise fits between molecules with specific shapes. One aspect of molecular
geometry that interests chemists is called the bond angle. It is the angle between
two bonds in a molecule. For example, the angle formed where two hydrogen
atoms link to an oxygen atom to form water (H2O) is about 104.5°.
Investigate
A water molecule can be studied in a
Cartesian plane. If we allow each unit on the
plane to represent 10"11 metres and place
the oxygen atom at the origin, then the
hydrogen atoms are located symmetrically at
about (7.59, 5.88) and (7.59, 5.88). The
bond angle formed at the oxygen atom is
•; rr
6 = 180  2 x tan'(yg) = 104.5°.
H
y
Nl
Y
/
642
X
2 4 6
46
Can you explain why this calculation is correct?
Nitrogen trioxide (NO3~) is an example of a trigonal planar
molecule. It consists of four atoms in a plane: three oxygen atoms
surrounding and individually bonding to a single nitrogen atom.
Because there are three identical atoms surrounding the nitrogen
atom, the three are evenly spaced around a circle. The bond angle
for each of the three bonds is, therefore, 360 s 3 = 120°.
DISCUSSION QUESTIONS
1. If the distance between the nitrogen atom and each oxygen atom in NO3" is
1.22 x 1010 metres, what is one way to assign planar coordinates to the
atoms?
2. Formaldehyde (H2CO) is a trigonal planar molecule with
the carbon in the centre. The bond between the carbon
and the oxygen is shorter than the bond between the
carbon and either one of the hydrogen atoms. Which is
likely to be smaller, the OCH bond angle or the HCH
O
'
x't"'ss
H
H
Formaldehyde
bond angle?
3. Can threeatom molecules always be studied in a plane? Can fouratom
molecules always be studied in a plane? What about molecules with more
than four atoms? ®
42
CHAPTER 2
48.
In this chapter, we establish principles that allow the use of algebraic methods in the
study of vectors. The application of algebra to problems in geometry first became
possible in 1637, when Descartes introduced the concept of a coordinate system.
A line is a geometrical object. How is a coordinate system
o
for a line constructed? First, choose an arbitrary point on the
line as a reference point, or origin. Next, associate with each
origin
point P on the line a real number a. How? Let the sign of a
indicate which side of the origin P is on, and let the magnitude of a represent the
distance from the origin to P. The result is known as the real number line, and a is
called the coordinate of P.
The correspondence between points on the line and real numbers is complete in
this sense: each point on the line has a different real number as its coordinate, and
every real number corresponds to one and only one point on the line.
Now let m be a vector on this line. Move the vector until its
p
H
initial point is at the origin. Its endpoint will fall on some
origin
point P with coordinate a. The coordinate a contains
everything you need to describe the vector u.
The absolute value
I a I is the magnitude of u, and the sign of a tells you its
direction.
We have now established the connection between the coordinates of a point and a
geometrical vector on a line. This amounts to an algebraic representation of a
geometrical vector. It is the first step in the development of algebraic methods to
handle vector problems.
A line is onedimensional. A plane has two dimensions.
But the same process leads to an algebraic representation
of a vector in a plane. The Cartesian coordinate system
b 
,P(.a, b)
for a plane is constructed from two real number lines—
the ,vaxis and the vaxis—placed at right angles in the
plane. The axes are oriented so that a counterclockwise
rotation about the origin carries the positive .vaxis into the positive .vaxis. Any
point P in the plane is identified by an ordered pair of real numbers (a, b), which
are its coordinates.
Let u be a vector in the plane. Move it until its initial point
is at the origin. Its endpoint will fall on some point P with
b)
coordinates (a, b). The magnitude of u can be determined
from («, b) using the Pythagorean Theorem. The direction
2.1 ALGEBRAIC VECTORS
43
49.
of u can be expressed in terms of the angle 9 between u and the positive jcaxis.
We can observe that, just as in the case of a line, the magnitude and direction of h
are determined entirely by the coordinates of P. Nothing else is needed.
Therefore, the ordered pair (a, b) is a valid representation of the vector h.
Any vector urn a plane can be written as an ordered pair (a, b), where its
magnitude I u  and direction 8 are given by the equations
and
8 = tan~'l—
with 8 measured counterclockwise from the positive jraxis to the line of the
vector. The formula above gives two values of 9, 0 s 9 < 360°. The actual
value depends on the quadrant in which P(a, b) lies.
The ordered pair (a, b) is referred to as an algebraic vector. The values of a and b
are the .v and jcomponents of the vector.
It is important to remember that the ordered pair (a, b) can be interpreted in two
different ways: it can represent either a point with coordinates a and b, or a vector
with components a and b. The context of a problem will tell you whether (a, b)
represents a point or a vector.
EXAMPLE 1
The position vector of a point P is the vector OP from the origin to the point.
Draw the position vector of the point P(3, 7), express it in ordered pair notation,
and determine its magnitude and direction.
Solution
The point P(—3, 7) is in the second quadrant. The position vector of P is
OP = (3, 7). The magnitude and direction of OP are
calculated as follows:
OP2 = (3)2 + (7)2
[b~P = V58
,
tan 6 = ^3
8= l>
Thus, the magnitude of OP is V58. Its direction makes an
angle of approximately 113° with the positive .vaxis.
Another notation commonly used to describe algebraic
vectors in a plane employs unit vectors. Define the vectors
/ =(1,0) andy = (0, 1). These are unit vectors that point in
the direction of the positive .iaxis and positive vaxis,
respectively.
44
CHAPTER 2
_3
50.
As you can see in the diagram, the position vector of
point P(a, b), and thus any vector it in the plane, can
a, b)
be expressed as the vector sum of scalar multiples of ;
and j.
i".
ai
Ordered pair notation and unit vector notation are equivalent. Any algebraic
vector can be written in either form:
u = 7)P = (a, b) or u =~OP = m + bj
EXAMPLE 2
Express the position vector of each of the points shown in the
y
diagram as an ordered pair and in unit vector notation.
fi(0, 7)
0(3. 3)
Solution
OP= (6, 2)
= 6/
2]
OQ = (3, 3)
= 3/
OR = (0, 7)
+ 37
P{6, 2)
= 7/
A coordinate system for threedimensional space is formed in much the same
way as a coordinate system for a twodimensional plane. Some point in space is
chosen as the origin. Through the origin, three mutually perpendicular number
lines are drawn, called the .Yaxis, the yaxis, and the axis. Each point in
space corresponds to an ordered triple of real numbers (a, b, c), which are its
coordinates on the three axes.
There are two different ways to choose the positive
directions of the axes. As a rule, mathematicians use a
righthanded coordinate system. If you could grasp the
caxis of a righthanded system with your right hand,
pointing your thumb in the direction of the positive zaxis,
your fingers should curl from the positive .Yaxis toward
the positive yaxis. A lefthanded system would have the
positive yaxis oriented in the opposite
direction.
A plane in space that contains two of
the coordinate axes is known as a
coordinate plane. The plane containing the
x and yaxes, for instance, is called the
Ayplane. The other two coordinate planes
are named similarly. A point such as
(—4, 0, I), which has a ycoordinate of 0.
lies in the .vzplane.
2.1 ALGEBRAIC VECTORS
45
51.
To plot a point P(a, b, c) in space, move a units from the origin in the x direction,
then b units in the y direction, and then c units in the z direction. Be sure each
move is made along a line parallel to the corresponding axis. Drawing a rectangular
box will help you to see the threedimensional aspect of such diagrams.
Just as in two dimensions, any vector in space can be placed with its initial point
at the origin. Its tip will then fall on some point P with coordinates {a, b, c), from
which its magnitude and direction can be determined. The ordered triple (a, b, c),
therefore, represents an algebraic vector in three dimensions. Alternatively, this
vector could be expressed in terms of unit vectors /,_/', and k, where / =(1,0, 0),
} = (0, 1,0), and A: = (0,0, 1).
Any vector u in threedimensional space can be written
as an ordered triple, u = OP — (a, b, c),
or in terms of unit vectors, u = OP = at + bj + ck.
Its magnitude is given by I«I = Va2 + b2 + c2.
EXAMPLE 3
Locate the point P, sketch the position vector OP in three dimensions, and
calculate its magnitude.
a. P(5, 7, 2)
b. ~OP = 3? + 5/ 4 it
Solution
b.
a.
. 5. A)
OP = V(5)2 + (7)2 + (2)2
= V78
46
CHAPTER 2
I OP I = V(3)2 + (5)2 + (4)2
= V50
52.
In two dimensions, we can describe the direction of a vector by a single angle. In
three dimensions, we use three angles, called direction angles.
The direction angles of a vector (a, b, c) are the angles a, P, and y that
the vector makes with the positive x,y, and zaxes, respectively, where
0° < oc. (J, y < 180°.
In this context, the components a, b, and c of the vector u are referred to as
direction numbers.
z
In the given diagram, the direction angles are all acute
_
angles. We can see the right triangle that relates m,
the direction number c, and the direction angle y,
from which it follows that cos y = t4t.
u
c
u/
V
a.'
y
The other direction numbers and angles are related in
the same way.
The direction cosines of a vector are the cosines of the direction angles
a, (3 and y, where
cos a = t=. , cos B = p.T and cos y = .=
Note that if you divide a vector (a, b, c) by its magnitude «, you create a unit
vector with components [Srr, rtr, r&A which is exactly (cos a, cos B, cos y).
HHI
IMI
IU '
Thus, the direction cosines are the components of a unit vector. Consequently,
cos2 a + cos2 p + cos2 y = 1.
It follows from this that the direction cosines, and hence the direction angles, are
not all independent. From any two of them you can find the third.
EXAMPLE 4
Find the direction cosines and the direction angles of the vector h = (0, 5, 3).
Solution
The magnitude of m is V(0)2 + (5)2 + (3)2 = V34.
The direction cosines and angles are therefore
cos a = tt=.
V34'
a = 90°
y= 121°
2.1 ALGEBRAIC VECTORS
47
53.
This vector is perpendicular to the xaxis, and is, therefore, parallel to the
yzplane.
EXAMPLE 5
A vector u makes angles of_60° and 105°, respectively, with the x and yaxes.
What is the angle between u and the zaxis?
Solution
cos2 60° + cos2 105° + cos2 y = 1
cos y = ± VI  cos2 60°  cos2 105°
Y= 34° or 146°
The angle between it and the zaxis is 34° or 146°, so there are two possible
vectors.
Exercise 2.1
Part A
1. What is the difference between an algebraic vector and a geometric vector?
2. Rewrite each of the following vectors in the form ai + bj.
a. (5,2)
b. (0,6)
c. (1,6)
3. Rewrite each of the following vectors as an ordered pair.
a. 2/+/
b. 3/
c. 5/ 5/
4. Rewrite each of the following vectors in the form ai + bj + ck.
a. (2,1,1)
b. (3,4,3)
c.
(0,4,1)
d. (2,0,7)
5. Rewrite each of the following vectors as an ordered triple,
a. 3/  8/ + k
b. 2/  2j  5*
c. 2/ + 6*
d. 4? + 9/
6. Express each of the following vectors as an algebraic vector in the fonn
(a, b).
a. m = 12,8= 135°
c. »v = 16, 6= 190°
48
CHAPTER 2
b.  v = 36, 0 = 330°
d. IxI = 13, 6 = 270°
54.
7. Express each of the following vectors as a geometric vector by stating its
magnitude and direction.
a. Tt = (6V3. 6)
b. v = {4V3, 12)
c. w = (4, 3)
d. .v = (0, 8)
8. What vector is represented in each of the following diagrams?
a.
„
I
b.
I
I
!
I
I
I
d.
c.
I I
I
I
I
■
z
I I
e.
z>' V
9. For each of the following, draw the .vaxis, yaxis, and caxis. and accurately
plot the points.
/U3,0,0)
5(0.2,0)
C(0,0,2)
D(3,2,0)
£(3,0,2)
f(0.2.3)
C(2,0,3)
//(0,3,2)
Part B
10. Describe where each of the following sets of points is located.
a. (0, 0, 6), (0, 0, 3), (0, 0,4)
b. (0, 2. 8), (0, 8, 2), (0, 2, 2)
c. (3. 0. 3). (3, 0. 5), (3. 0, 5)
d. (1, 2, 0), (0,4, 0), (5, 6, 0)
e. (1, 3, 2), (I, 3, 6), (1, 3, 11)
f.
(2. 2. 2). (3. 3. 3), (8, 8, 8)
2.1 ALGEBRAIC VECTORS
49
55.
11. Where are the following general points located?
a. /(.v,y, 0)
b. J3(jc, 0, 0)
c.
d. D(0, 0, z)
C(0,y, z)
e. E(x,Q,z)
f.
F(0,y,0)
12. For each of the following, draw the .raxis, yaxis, and zaxis and accurately
draw the position vectors.
a. M(6,4, 2)
b. N(3,5, 3)
c. P(2,3.7)
d. G(4,9,5)
e. /?(5, 5, 1)
f.
T(6, 1,8)
13. Find the magnitude and the direction of the following vectors,
a. 0E = (1,7)
b. OF =(0,6)
c. OG = (9, 12)
d. ~OH = (j, ^)
e. OJ = (^, ^)
f. ~OK = (V6, 0)
14. Find the magnitude of the following vectors.
a. (12,4,6)
b. (8,27,21)
c. (^^^f)
d. (V2.2V3, V2)
15. Can the sum of two unit vectors be a unit vector? Explain. Can the difference?
16. a. Calculate a when a = (2, 3, 2).
b. Find T=T« Is it a unit vector?
17. a. Find the magnitude of the vector v = 2/ — 3/ — 6it.
b. Find a unit vector in the direction of v.
18. If v = (3, 4, 12), find a unit vector in the direction opposite to w
19. Show that any unit vector in two dimensions can be written as (cos 6, sin 6),
where 0 is the angle between the vector and the .vaxis.
20. Reposition each of the following vectors so that its initial point is at the
origin, and determine its components.
a.
y
I I I V I I I
50
CHAP1ER 2
I
b.
I
I I
I
I
I I
i i I i i i i r
56.
c.
d.
++4
X
"•
21. Draw a diagram of a vector it = (a, b, c) that illustrates the relationship
between
a. u, a, and cos a (a acute)
b. u, b, and cos P (P obtuse)
22. The direction angles of a vector are all equal. Find the direction angles to the
nearest degree.
PartC
23. Prove that the magnitude of the vector OP = (a, b, c) is given by
O? = Va2 + 62 + c2.
24. Give a geometrical interpretation of the vector « = (4, 2, —5, 2).
Make a reasonable conjecture about its magnitude.
2.1 ALGEBRAIC VECTORS
51
57.
Section 2.2 —
As we saw in Section 2.1, all vectors can be expressed in terms of the unit vectors
i and/ in two dimensions, or i,j, and k in three dimensions, or, equivalently, in
terms of ordered pairs or triplets. Vectors such as i,j, and k, which have been
chosen to play this special role, are termed basis vectors. They form a basis for
the two or threedimensional spaces in which vectors exist. In Example 1. we
establish the uniqueness of the algebraic representation of a vector in terms of
these basis vectors.
EXAMPLE 1
Prove that the representation of a twodimensional algebraic vector in terms of its
x and ycomponents is unique.
Solution
Using the method of proof by contradiction, we begin by assuming that the vector
u can be written in terms of components in two different ways:
u = aj + bj and u = a2i + b2j
Since they represent the same vector, these expressions must be equal.
a' + bJ = a2' + bij
Some rearrangement produces the equations
fli a2i = ~bj + b2j
(a,  a2)i = (~bl + b2)j
The last equation states that a scalar multiple of/ equals a scalar multiple ofy.
But this cannot be true. The unit vectors / and j have different directions, and no
multiplication by a scalar can make the vectors equal. The only possible way the
equation can be valid is if the coefficients of / andy are zero, that is, a, = a2 and
bx = b2, which means that the two representations of the vector m are not different
after all.
The uniqueness of algebraic vectors leads to a fundamental statement about the
equality of algebraic vectors.
Two algebraic vectors are equal if and only if their respective
Cartesian components are equal.
52
CHAPTER 2
I
58.
Since all vectors can be expressed in terms of the basis vectors i,j, and k, all
the rules of vector algebra discussed in Chapter 1 apply to algebraic vectors.
In two dimensions, for instance, scalar multiplication of a vector and addition
of two vectors, written in both unit vector and ordered pair notation, look like
this:
k(ai + bj) = kai + kbj
scalar multiplication
or
k(a. b) = (to, kb)
+ bj) + (a2i + b2j) = («
vector addition
b) + (a2, b2) = (« + a2,
EXAMPLE 2
b2)j
a2)i +
or
+ b2)
If u = (5, 7) and v = (2. 3), find w = 6»  4r.
Solution
In ordered pair notation
In unit vector notation
w = 6u — 4v
w = 6h — 4v
= 6(5,7)4(2, 3)
= 30/ 42/ +8/
= (38, 54)
EXAMPLE 3
= 6(5/'  Ij)  4(2/ + 3/)
= (30, 42) + (8. 12)
= 38/  54/
 12/
Using vectors, demonstrate that the three points A(5. 1), B( 3. 4), and
C( 13. —6) are collinear.
Solution
The three points will be collinear if the vectors AB and BC have the same
direction, or the opposite direction.
A~B = (&, 5)
~BC = (16, 10)
ThenBC= 2AB
AB and BC have the opposite direction, so the points A. B, and C must be
collinear.
EXAMPLE 4
If A(, 5. 2) and fl(3. 4, 4) are opposite vertices of parallelogram OAPB and O
is the origin, find the coordinates of P.
2.2 OPERATIONS WITH ALGEBRAIC VECTORS
53
59.
Solution
But
SP = G4 = (1,5, 2)
Then OP = (3,4,4) +(I,5,2)
= (2,1,6)
Therefore, point P has coordinates (—2, —1,6).
Exercise 2.2
Part A
1. In two dimensions, the unit vectors / and./ have been chosen as the basis
vectors in terms of which all other vectors in the plane are expressed.
a. Consider the merits of this choice as opposed to using vectors that do not
have unit magnitude.
b. Consider the merits of this choice as opposed to using vectors that are not
perpendicular.
2. Find a single vector equivalent to each expression below,
a. (2,4)+ (1,7)
b.
5(1,4)
c. 0(4,5)
d. (6,0)+ 7(1,1)
e. (2,1,3)+ (2, 1,3)
f.
g. (4,1,3)(2, 1,3)
h. 2(l,l,3) + 3(2,3,1)
i. 2(0, 1,0)+ 5(0, 0,1)
j. {(4,6, 8) + (4,6, 8)
k. 5(0, 2, 4)  4(3, 8,0)
1.
2(1,1,4)
2(3, 2,4) + 5(3, 2, 8)
3. Simplify each of the following expressions.
a. (2/ + 3j) + 4(/  j)
b. 3(?  2j + 3k)  3(f + 4/  3A)
c. 3(i  A)  (2i + k)
d. 5(9/  Ij)  5(9/ + 7Jt)
4. Given a = (2, 1, 4), b = (3, 8, 6), and c = (4, 2, 1), find a single vector
equivalent to each of the following expressions.
a. 2a  b
b. a  £
c. 3a  ~b  2c
d. a + b + 2c
e. 2a + ~b  c
f. 4a  2b + c
5. Given x = 2» — j + k and y = 2/ + 4k, express each quantity in terms of
i,j, and k.
54
CHAPTER 2
a.
3x + y
b. .v + y
c. x — y
d. y  .v
60.
6. If a — 3/ + 2/  A and b = —2/ +_/, calculate each magnitude,
a. a + b
b. a2>
c. 12a — 3A I
7. If £)(3, 4, 5) and E( —2, 1, 5) are points in space, calculate each expression
and state what it represents.
a. o6
d. D£
b. OE
e. ED
c. ~DE
f. !Z5
PartB
8. Using vectors, demonstrate that these points are collinear.
a. P(15, 10), Q(6, 4), and «(12, 8)
b. D(33, 5, 20), £(6, 4, 16), and F(9, 3, 12)
9. For each set of points A, B, C, and D, determine whether AB is parallel to CD
and whether AB* = cd.
a. A(2, 0), B(3, 6), C(4, 1), D(5, 5)
b. A(0, 1,0), B(4, 0, 1), C(5, 1, 2), D(2, 3, 5)
c. A(2, 4, 6), BO, 4, 1), C(4, 1, 3), D(5, 1,2)
10. If PQRS is a parallelogram in a plane, where P is (4, 2), 0. is (—6, 1), and 5 is
(—3, —4), find the coordinates of/?.
11. If three vertices of a parallelogram in a plane are (5, 3), (5, 2), and (7, 8),
determine all the possible coordinates of the fourth vertex.
12. If Q4, OB, and 7>C are three edges of a parallelepiped where O is (0, 0, 0),
A is (2, 4, 2), B is (3, 6, 1), and C is (4, 0,1), find the coordinates of the
other vertices of the parallelepiped.
13. A line segment has endpoints with position vectors OA{ and OA 2.
The midpoint of the line segment is the point with position vector
Find the position vector of the midpoint of the line segment from
a. A(5, 2)toB(13, 4)
b.
C(3, 0) to D(0,7)
c. £(6, 4, 2) to F(2, 8, 2)
d. G(0, 16, 5) to //(9, 7, 1)
14. a. Find x and v if 3(.t, 1)  2(2, y) = (2, 1).
b. Find .v, y, and z if 2(.v, 1, 4)  3(4, y, 6)  j(4, 2, z) = (0, 0, 0).
2.2 OPERATIONS WITH ALGEBRAIC VECTORS
55
61.
PartC
15. Find the components of the unit vector with direction opposite to that of the
vector from X(7, 4, 2) to Y( 1, 2, 1).
16. a. Find the point on thevaxis that is equidistant from the points (2, — 1, 1)
and(0, 1,3).
b. Find a point not on the yaxis that is equidistant from the points (2, — 1, 1)
and(0, 1,3).
17. a. Find the length of the median AM in the triangle ABC, for the points
A(2, j, 4), B(3, 4, 2), and 0(1, 3, 7).
b. Find the distance from A to the centroid of the triangle.
18. The centroid of the n points with position vectors OAX, 0A2, ..., 0An is the
point with position vector
— ^ O4, + Q42 + ... + 0A,,
n
Find the centroid of each of the following sets of points.
a. A(l,2),fl(4,l).C(2,2)
b. /(I,O,O),J(O, l,O),tf(O,O, 1)
c. A,(3, 1), A2( 1, 1), A3(7, 0), A4{4, 4)
d. C(0, 0, 0), /(1, 0, 0), 7(0, 1, 0), K(0, 0, 1)
19. The centre ofjnass ofthe masses m{, m2, ..., »in at the points with position
vectors OAX, OA2
OG =
OAIV respectively, is the point with position vector
m)0Ai + ... + mn0An
n0An
m2
In some kinds of problems, a collection of masses can be replaced by a single
large mass M = in + m, + ... + mn located at the centre of mass, for the
purposes of calculation. Calculate the centre of mass in each case.
a.
A mass of 2 units at (0, 0), a mass of 3 units at (4, 1), a mass of 5 units at
(— 1, —7), and a mass of 1 unit at (11, —9)
b. A mass of 1 unit at (1, 4, —1), a mass of 3 units at (2, 0, 1), and a mass
of 7 units at (1, 3. 10)
56
CHAPTER 2
62.
Certain applications of vectors in physics and geometry cannot be handled by
the operatons of vector addition and scalar multiplication alone. Other, more
sophisticated combinations of vectors are required. The dot product of two
vectors is one of these combinations.
The dot product of two vectors u and v is
n a v =
r[ cos
//
where i) is (lie ansjle between the two vectors.
Since the quantity ii v cos 0 on the right is the product of three scalars, the
dot product of two vectors is a scalar. For this reason, the dot product is also
called the scalar product.
EXAMPLE 1
Find the dot product of u and v in each of the following cases, where fl is the
angle between the vectors.
a. u =7, v = 12, 8 = 60°
b. « =20, v = 3.6 = ^
c. it =24, I v I = 9, fl = 34°
Solution
_
a. u • v = u vI cos 60"
b. u • v = it v cos ^
= 3OV3
= 42
c. u • v =  u I  v  cos 34°
= (24)(9)(0.8290)
s 179.1
EXAMPLE 2
Prove that two nonzero vectors // and v are perpendicular, if and only if it • v = 0.
Proof
The condition that it • v = 0 is sufficient. Nothing else is needed to guarantee that
the vectors are perpendicular, because
if
it • »• = 0
then
1171  v  cos 6 = 0
or
cos 6 = 0 (since the vectors are nonzero)
2.3 THE DOT PRODUCT OF TWO VECTORS
57
63.
Therefore,
6 = ± 90°,
which means that the vectors must be perpendicular.
The condition that u • v = 0 is necessary, because
if
« • v * 0
then
1771 v cose # 0,
which means that cos 6 cannot be zero.
Consequently, 8 cannot be 90°.
So the vectors are perpendicular only ifu • v = 0 .
The following properties of the dot product will be demonstrated in Exercise 2.3.
You can multiply by a scalar either
a(u • v) = (au) • v = i* • (av)
before or after taking the dot product.
You can expand a dot product of a vector
u • (v + w) = u • v + u • w
with the sum of two other vectors as you would in ordinary multiplication.
The dot product of a vector u with itself
u • u = «2
is the square of the magnitude of the vector.
Dot products of the basis vectors i,j, and k are of particular importance.
Because they are unit vectors,
Because they are perpendicular,
/ • / = I
/ »j = j • i =0
j •] = I
k•k= 1
j • k = ic»j = 0
ic»l = i'k = 0
The dot product is 1 if the vectors are the same, and 0 if they are different.
These results are used to work out the dot product of two algebraic vectors,
which, for vectors in space, proceeds in this manner:
If
u = uj + ujc + uzk and v = vxi + vvj + vjc
then u • v = (mv/ + uyj + «.£) • (vr/ + vyj + vjc)
= uxvx0 • /) + uxvy(i •]) + uxv.U • k)
+ UyVjj • /) + UyV/j •]) + uyvz(j • k)
+ uzvx(ic • i) + uzvy(k »j) + u.v.(k • k)
= uxvx() + uxvy(0) + uxv.(0)
+ UyVjjQ) + UyVy(l) + UyVJLO)
+ uzvx(0) + uz>y(Q) + m.vz(1)
UyVy + U.V
58
CHAPTER 2
64.
EXAMPLE 3
Find the dot product of u and v, where
a. h = (5, 2) and v = (3,4)
b. u = (1, 0,4) and v = (2, 5, 8)
Solution
h • v = (5, 2) • (3, 4)
if • v = (1, 0, 4) • (2, 5, 8)
= 15 + 8
EXAMPLE 4
= 2 + 0 + 32
= 7
= 30
Find the angle 8 between each of the following pairs of vectors.
a. « = (6, 5) and v = (1, 3)
b. h = (3, 1, 2) and v = (5, 4, 1)
Solution
Since u • v = u v cos 6,
then cos 8 = i'i ii .
I '< I I v I
(6. 5) ■•(1. 3)
1(6,5)1 (1, 3)1
(6WD + (5W3)
V(6)2 + ( 5)2V(l)2 + (3)2
9
VSTVIo
..8 s
= 0.3644
69°
(3,1. 2) • (5.4.1)
I (3. 1,2) 1(5,,4,1)1
(3K5) + (l)(4) + (2)(l)
V(3r + (I)2 + (2)2 V(5)2 + (4)2 + (I)2
21
V3
>
.. 8 = 150°
2.3 THE DOT PRODUCT OF TWO VECTORS
59
65.
Exercise 2.3
Part A
1. a. What is the dot product of two vectors if the angle between them is 0°?
90°? 180°?
b. What is the angle between two vectors if their dot product is positive?
negative? zero?
2. Calculate the dot product u • v, given the magnitudes of the two vectors and
the angle 8 between them.
a. ii =3, v =4,0 = 45°
c. u = 9, v =3,6 = ^f
b. u = 6,  r = 5, 8 = 60°
d. u = =, v =,8 = 90°
3. Examine each of the following pairs of vectors. State whether or not the
vectors are perpendicular, then sketch each pair, and find their dot product.
a. i/ = (4, 1), b = (1,4)
b. c = (5, 2), d = (5, 2)
c. p = (1,0), ? = (0,1)
d. u = (7, 8), v = (4, 7)
4. Find the dot product of each of the following pairs of vectors and state which
pairs are perpendicular.
a. a = (1,3, 4), b = (1,3, 2)
b. .v = (2, 2. 4), v = (4. 1,2)
c. m = (5, 0, 0), h = (0, 3, 0)
d. 7 = (0, 3, 4), 7 = (0, 3, 4)
e. h = (0, 5, 6), v = (7, 0, 1)
f.
c = (8, 11, 5),
d= (7. 11,13)
5. a. Find three vectors perpendicular to (2, 3).
b. How many unit vectors are perpendicular to a given vector in the .vyplane?
6. a. Find three noncollinear vectors perpendicular to (2, —3, 1).
b. How many unit vectors are perpendicular to a given vector in three
dimensions?
7. Calculate, to four decimal places, the cosine of the angle between each of the
following pairs of vectors.
a. a = (8, 9), b = (9, 8)
60
CHAPTER 2
b. c = (1,2, 3),rf = (4, 2,1)
66.
Part B
8. Determine the angle between the following vectors.
a. a = (X5).b = (4, 1)
b. c = (5. 6. 7). d = (2.3. 1)
c. i = (1,0, 0).m = (1, I, 1)
d. p = (2, 4, 5), q = (0,2,3)
9. Given a = (2. 3. 7) and b = (4. y, 14).
a.
for what value of y are the vectors collinear?
b. for what value of v are the vectors perpendicular?
10. Find any vector »r that is perpendicular to both u = 3j + 4k and v = 2i.
11. If the vectors a = (2. 3. 4) and b = (10, v, z) are perpendicular, how must y
and z be related?
12. For» = (1. 5. 8) and v = (1,3. 2), verify that
a. u * v = v ii
b. u • u =  u 2 and v * v = v p
c. (» + r)«(«r)= «2 v2
d. (h + v) • (it + v) = 17t  2 + lit • v +  v  2
e.
(2u) • v = it • (2v) = 2(» • v)
13. lf« = (2, 2, I). v = (3, 1,0), and w = (1,7,8). verify that
It • (l' + W) = II • V + If W.
14. Expand and simplify.
a. (4/  ))']
b. k • {] ~ 3A)
c. (/"  4k) • (i*  4k)
15. Expand and simplify.
a. (3fl + 4/>) • (5a + 6b)
b. (2a  b) • (2a + b)
16. Find (3a + b) • (2ft  4«), if o = /  3/ + k and b = 2i + 4/  5Jt.
17. Two vectors 2a + b and a  3b are perpendicular. Find the angle between a
andft.if o =2b.
18. Given a and 6 unit vectors.
a. if the angle between them is 60°, calculate (6a + b) • (a  2b)
b. if  a + b = V'X determine (2a  5b) • (b + 3a)
19. The vectors a = 3/  4j  k and b = 2i + 3j  6k are the diagonals of a
parallelogram. Show that this parallelogram is a rhombus, and determine the
lengths of the sides and the angles between the sides.
2.3 THE DOT PRODUCT OF TWO VECTORS
61
67.
20. a. If a and b are perpendicular, show that I a 12 + I £ 12 = I a + £ 2.
What is the usual name of this result?
b. If a and 6 are not perpendicular, and a  ~b = c, express  c 2 in terms of
a and 7). What is the usual name of this result?
PartC
21. If the dot product of a nnd_b is equal to the dot product of a and c, this does
not necessarily mean that b equals c. Show why this is so
a. by making an algebraic argument
b. by drawing a geometrical diagram
22. Find a unit vector that is parallel to the .vyplane and perpendicular to the
vector 4/  3/ + k.
23. Three vectors .v, v, and z satisfy x + y + z = 0. Calculate the value of
x• y + y • z + z • .v, if .v =2, y =3, and z =4.
24. A body diagonal of a cube is a line through the centre joining opposite
vertices. Find the angles between the body diagonals of a cube.
25. a. Under what conditions is (a + b) • (a  b) = 0? Give a geometrical
interpretation of the vectors a, b, a + b, and a — b.
b. Use the dot product to show that two vectors, which satisfy the equation
_[h + r I = I u — v , must be perpendicular. How is the figure defined by
it and v related to the figure defined by a and b of part a?
26. Prove that a#/> ^ a b. When does equality hold? Express this
inequality in tenns of components for vectors in two dimensions and for
vectors in three dimensions. (This is known as the CauchySchwarz
Inequality.)
62
CHAPTER 2
68.
We have already defined the dot product of two vectors, which gives a scalar quantity.
In this section we introduce a new product called the cross product or vector
product. The cross product of two vectors a and b is a vector that is perpendicular
to both a and b. Hence, this cross product is defined only in threedimensional
space. The cross product is useful in many geometric and physical problems in
threedimensional space; it is used to help define torque and angular velocity in
statics and dynamics, and it is also used in electromagnetic theory. We will use it
to find vectors perpendicular to two given vectors.
Let a = (fl. o>, ay) and b = (b, b>, by) be two
in threedimensional space. Let us find all the vectors
a x b
v = (.v, y, z) that are perpendicular to both a and b.
These vectors satisfy both
a • v = 0 and b • v = 0.
Hence,
*a
fl,.v + Ot}' + ayz = 0
'
Z? j.v + btv + byz = 0

We solve these two equations for.v. y, and z. Multiply equation '1 by /73. and
equation I by a3 to obtain
aybfX + ajb^y + n^b^z = 0
•'
Now eliminate z by subtracting equation 3 from equation 4 to obtain
(o3/j — a^b^x + (a3b2  (ijb^y — 0
This is equivalent to
Using a similar procedure, we eliminate x from the original equations to obtain
V;
Let v = k{a,b —
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